Binomial Trials And Binomial Distribution

Q: What is the binomial distribution formula?

If X counts successes in n independent trials with success probability p, then P(X = r) = nCr p^r q^(n - r) with q = 1 - p. The nCr counts the orders of the successes, and p^r q^(n - r) is the probability of each order.

Q: What are the mean and variance of a binomial distribution?

For X ~ B(n, p), the mean is np and the variance is npq, so the standard deviation is the square root of npq. Because q is less than 1, the variance is always smaller than the mean. Many questions give the mean and variance and ask you to recover n and p.

Q: How do you find the most probable number of successes?

Compare consecutive terms using P(X = r + 1) / P(X = r) = ((n - r)/(r + 1)) (p/q). The most likely value is the integer part of (n + 1)p. If (n + 1)p is an integer, that value and the one just below it are equally likely.

Q: What is the expected value of a random variable?

The expected value E(X) = sum of x_i p_i is the long-run average of X over many repetitions, the balance point of its distribution. It need not be a value X can take: a fair die has E(X) = 3.5. For a game, a positive expected gain means you win on average.

Q: Is binomial distribution in the JEE Main 2026 syllabus?

The revised JEE Main syllabus lists probability distribution of a random variate but no longer names Bernoulli trials and the binomial distribution separately. Repeated independent trial questions can still be solved with the multiplication theorem and combinations, so the binomial formula remains a useful and time-saving tool.

MathsProbabilityFor JEE aspirants

A binomial distribution gives the probability of exactly $r$ successes in $n$ independent trials, each with the same success probability $p$ : $P (X = r) =^{n} C_{r} p^{r} q^{n - r}$ with $q = 1 - p$ . Such repeated success-or-failure trials are called binomial (Bernoulli) trials. This binomial distribution page also covers random variables, probability distributions, mean and variance, the results $n p$ and $n pq$ , the most probable value and first-success problems, with solved examples for JEE Main and JEE Advanced.

On this page1Random variable2Mean and variance3Bernoulli trials4Binomial formula5Mean np, variance npq6Strategy7Solved examples

Key Formulas - Quick Reference

Probability distribution: $p_{i} \geq 0$ and $\sum p_{i} = 1$
★ Must learnMean (expectation): $μ = E (X) = \sum x_{i} p_{i}$
★ Must learnVariance: $σ^{2} = \sum x_{i}^{2} p_{i} - μ^{2}$ ; standard deviation $σ = σ^{2}$
$E (a X + b) = a μ + b$ and $Var (a X + b) = a^{2} Var (X)$
★ Must learnBinomial: $P (X = r) =^{n} C_{r} p^{r} q^{n - r}$ , $r = 0, 1, \dots, n$ , $p + q = 1$
At least one success: $1 - q^{n}$ ; first success on the $r$ -th trial: $q^{r - 1} p$
★ Must learnBinomial mean $= n p$ , variance $= n pq$ , so variance $<$ mean
Recurrence: $\frac{P ( X = r + 1 )}{P ( X = r )} = \frac{n - r}{r + 1} \cdot \frac{p}{q}$
Most probable value (mode): integer part of $(n + 1) p$ ; if $(n + 1) p$ is an integer, both $(n + 1) p$ and $(n + 1) p - 1$

Syllabus note: JEE Main 2026 lists "probability distribution of a random variate" but no longer names Bernoulli trials and the binomial distribution. JEE Advanced 2026 lists random variables with their mean and variance. Sections 1 and 2 are core for both exams. Repeated independent trials can always be solved with the multiplication theorem and counting, so learn the binomial results as high-value shortcuts.

1. Random Variable and Probability Distribution

A random variable $X$ is a real-valued function defined on the sample space of a random experiment. It attaches a number to every outcome, such as the number of heads, the sum on two dice, or a prize amount.

The probability distribution of $X$ lists every value $x_{i}$ with its probability $p_{i} = P (X = x_{i})$ . It shows how the total probability 1 is shared among the values. A valid distribution needs $p_{i} \geq 0$ for every $i$ and $\sum p_{i} = 1$ .

Figure 1:

X

= number of heads maps

H H \to 2

H T, T H \to 1

T T \to 0

. Adding the probabilities of the outcomes sent to each value gives the distribution, which totals

1

Several outcomes can map to the same value (both $H T$ and $T H$ give $X = 1$ ), so $P (X = x_{i})$ is the sum of the probabilities of all outcomes sent to $x_{i}$ .

Exam Trick

Before finding a mean or variance, always confirm $\sum p_{i} = 1$ . Questions often hide an unknown $k$ in the table; the quadratic for $k$ usually gives one negative root, which must be rejected. Also reject any root that makes some $p_{i} > 1$ .

2. Mean and Variance of a Random Variable

For a random variable $X$ taking values $x_{1}, x_{2}, \dots, x_{n}$ with probabilities $p_{1}, p_{2}, \dots, p_{n}$ :

μ = E (X) = i = 1 \sum n x_{i} p_{i}

σ^{2} = Var (X) = i = 1 \sum n (x_{i} - μ)^{2} p_{i} = i = 1 \sum n x_{i}^{2} p_{i} - μ^{2}

The mean is the long-run average value of $X$ over many repetitions, which is why it is also called the expected value. It need not be a value $X$ can actually take. The variance measures how widely the values spread around the mean.

Figure 2: Treat the probabilities as weights on a beam: it balances at the mean

μ = 3

. Variance

= \sum x^{2} p - μ^{2} = 0.6

Two distributions can share a mean and still differ completely in spread. Figure 3 moves the outer values twice as far from the mean: the mean stays 3 but the variance becomes four times as large.

Figure 3: Both distributions balance at

μ = 3

, but the right one is spread twice as far, so its variance is

4 \times 0.6 = 2.4

. The mean says where; the variance says how spread.

2.1 Linear change of a random variable

$E (a X + b) = a E (X) + b$ : shifting and scaling act on the mean directly.
$Var (a X + b) = a^{2} Var (X)$ : adding a constant $b$ does not change the spread, and scaling by $a$ multiplies it by $a^{2}$ .
$σ_{a X + b} = ∣ a ∣ σ_{X}$ .

Key idea

Mean = balance point

\sum x p

; variance =

\sum x^{2} p - μ^{2}

measures spread. Check

\sum p = 1

first.

Quick Recall: tap to check

X

takes values 1, 2, 3 with probabilities

k

2 k

3 k

. Find

k

6 k = 1

, so

k = \frac{1}{6}

E (X) = 4

and

Var (X) = 3

, find

E (2 X - 1)

and

Var (2 X - 1)

7

and

4 \times 3 = 12

Can the mean of a die score,

3.5

, ever be rolled?

No. The expected value need not be a possible value.

3. Binomial Trials (Bernoulli Trials)

Consider a random experiment whose result is classified as success or failure. The trials are called binomial (Bernoulli) trials if:

there is a fixed, finite number of trials $n$ ;
each trial has exactly two outcomes, success or failure;
the probability of success $p$ (and of failure $q = 1 - p$ ) is the same in every trial;
the trials are independent of each other.

Tossing a fair coin several times is the standard example: each toss gives a success (say a head) or a failure, with $p = q = \frac{1}{2}$ every time.

Binomial: draws WITH replacement

The ball goes back, so $p$ stays the same and the draws are independent.

3 cards with replacement: number of aces $\sim B (3, \frac{1}{13})$ .

Not binomial: draws WITHOUT replacement

$p$ changes from draw to draw, so the trials are dependent.

Count with $^{n} C_{r}$ or use the multiplication theorem (Solved Example 15).

Figure 4: Every sequence with 2 successes has probability

p^{2} q

, and there are

^{3} C_{2} = 3

of them, so

P (X = 2) = 3 p^{2} q

4. Binomial Distribution Formula

Repeat a binomial experiment $n$ times with success probability $p$ and failure probability $q$ in each trial. We want the probability of exactly $r$ successes.

The positions of the $r$ successes among the $n$ trials can be chosen in $^{n} C_{r}$ ways.
Any one such sequence of $r$ successes and $n - r$ failures has probability $p^{r} q^{n - r}$ , by independence.
The sequences are mutually exclusive, so their probabilities add.

If $X$ is the number of successes in $n$ binomial trials, then

P (X = r) =^{n} C_{r} p^{r} q^{n - r}, r = 0, 1, 2, \dots, n

This is the binomial distribution, written $X \sim B (n, p)$ . The probabilities are the successive terms of the binomial expansion

(q + p)^{n} =^{n} C_{0} q^{n} +^{n} C_{1} p q^{n - 1} +^{n} C_{2} p^{2} q^{n - 2} + \dots +^{n} C_{n} p^{n} = 1

Figure 5:

B (7, \frac{1}{2})

is symmetric, so

P (X \geq 4) = P (X \leq 3) = \frac{1}{2}

without any addition. The bar heights are the binomial coefficients

^{7} C_{r}

Exam Trick

When $p = q = \frac{1}{2}$ , the distribution is symmetric: $P (X = r) = P (X = n - r)$ . For odd $n$ , $P (X \geq \frac{n + 1}{2}) = \frac{1}{2}$ immediately, with no terms to add.

4.1 Standard results

Probability of at most $r$ successes $= \sum_{k = 0}^{r}^{n} C_{k} p^{k} q^{n - k}$ .
Probability of at least $r$ successes $= \sum_{k = r}^{n}^{n} C_{k} p^{k} q^{n - k}$ .
Probability of at least one success $= 1 - q^{n}$ .
Probability that the first success comes on the $r$ -th trial $= q^{r - 1} p$ .

Figure 6: The first success on trial

r

needs

r - 1

failures first, so

P = q^{r - 1} p

. For the first six on the 4th roll:

(\frac{5}{6})^{3} \cdot \frac{1}{6} = \frac{125}{1296}

Figure 7:

1 - q^{n}

climbs towards

1

but never reaches it. The first

n

with

1 - (\frac{5}{6})^{n} \geq 0.9

n = 13

(Solved Example 8).

4.2 Most probable number of successes

Dividing consecutive terms gives the recurrence

\frac{P ( X = r + 1 )}{P ( X = r )} = \frac{n - r}{r + 1} \cdot \frac{p}{q}

The probabilities rise while this ratio exceeds 1 and fall after it drops below 1. The ratio is greater than 1 exactly when $r < (n + 1) p - 1$ , so the most probable value (mode) is the integer part of $(n + 1) p$ . If $(n + 1) p$ is itself an integer, both $(n + 1) p$ and $(n + 1) p - 1$ are equally likely (Figure 8).

Figure 8: For

B (9, 0.4)

(n + 1) p = 4

is an integer, so

P (X = 3) = P (X = 4) \approx 0.251

: two equal peaks at

4

and

4 - 1 = 3

Key idea

Binomial = number of arrangements

^{n} C_{r}

times the probability

p^{r} q^{n - r}

of each one. Use

1 - q^{n}

for "at least one".

Quick Recall: tap to check

A fair coin is tossed 6 times. Find P(exactly 4 heads).

^{6} C_{4} (\frac{1}{2})^{6} = \frac{15}{64}

Find the most probable number of heads in 10 tosses of a fair coin.

(n + 1) p = 5.5

, so

5

A die is rolled until a six appears. Find P(first six on the 2nd roll).

\frac{5}{6} \cdot \frac{1}{6} = \frac{5}{36}

5. Mean and Variance of the Binomial Distribution

If $X \sim B (n, p)$ , then

Mean = μ = n p Variance = σ^{2} = n pq Standard deviation = n pq

Since $q < 1$ , the variance of a binomial distribution is always less than its mean.

Why $n p$ ? Each trial contributes 1 success with probability $p$ and 0 with probability $q$ , so its expected contribution is $p$ . Adding over $n$ trials gives $n p$ . Each trial has variance $p - p^{2} = pq$ , and variances of independent trials add, giving $n pq$ (Figure 10).

Figure 9: For

n = 10

p = 0.2

skews right,

p = 0.5

is symmetric,

p = 0.8

skews left. The peak sits near the mean

n p

, and

p

and

1 - p

give mirror images with the same variance.

Figure 10: For every

0 < p < 1

pq < p

, so

n pq < n p

: variance is always less than mean. Also

n pq \leq \frac{n}{4}

, with equality at

p = \frac{1}{2}

JEE Advanced

Mean from the formula. Using $r^{n} C_{r} = n^{n - 1} C_{r - 1}$ :

E (X) = r = 1 \sum n r^{n} C_{r} p^{r} q^{n - r} = n p r = 1 \sum n^{n - 1} C_{r - 1} p^{r - 1} q^{n - r} = n p (q + p)^{n - 1} = n p

A similar step with $r (r - 1)^{n} C_{r} = n (n - 1)^{n - 2} C_{r - 2}$ gives $E (X^{2}) = n (n - 1) p^{2} + n p$ , so $Var (X) = n pq$ .

Waiting time. If $Y$ is the trial on which the first success occurs, $E (Y) = \sum r q^{r - 1} p = \frac{p}{( 1 - q ) ^{2}} = \frac{1}{p}$ : on average you wait 6 throws for a six.

Exam Trick

Given the mean and variance, divide: $q = \frac{variance}{mean}$ , then $p = 1 - q$ and $n = \frac{mean}{p}$ . If the ratio is 1 or more, no binomial distribution exists.

Key idea

Mean

n p

, variance

n pq

, and variance is always less than mean. The ratio variance/mean is

q

6. How to Solve Binomial Problems

Check the four conditions: fixed $n$ , two outcomes, same $p$ , independent trials.
Define success clearly and write $n$ , $p$ , $q$ .
Translate the question: exactly $r$ , at least $r$ , at most $r$ , at least one, first success.
Use the complement whenever it has fewer terms, then simplify with the common factor $\frac{1}{2 ^{n}}$ or $q^{n}$ .

Figure 11: Check the four binomial conditions first; then the wording of the question picks the formula.

Figure 12: Revision map of the page, from a general random variable to the binomial distribution.

7. Solved Examples

7.1 Random variables, mean and variance

Solved Example 1

Two fair coins are tossed. Let

X

be the number of heads. Write the probability distribution of

X

Solution:

$S = {H H, H T, T H, T T}$ , so $X$ takes the values 0, 1, 2 (Figure 1).

$X$	0	1	2
$P (X)$	$\frac{1}{4}$	$\frac{2}{4}$	$\frac{1}{4}$

Check: $\frac{1}{4} + \frac{1}{2} + \frac{1}{4} = 1$ .

Solved Example 2

A random variable

X

has the distribution below. Find (i)

k

(ii)

P (X < 3)

(iii)

P (X > 6)

(iv)

P (0 < X < 3)

$X$	0	1	2	3	4	5	6	7
$P (X)$	0	$k$	$2 k$	$2 k$	$3 k$	$k^{2}$	$2 k^{2}$	$7 k^{2} + k$

Solution:

(i) The probabilities must add to 1: $10 k^{2} + 9 k = 1$ , so $10 k^{2} + 9 k - 1 = 0$ , giving $(10 k - 1) (k + 1) = 0$ . A probability cannot be negative, so $k = \frac{1}{10}$ .

(ii) $P (X < 3) = 0 + k + 2 k = 3 k = \frac{3}{10}$

(iii) $P (X > 6) = P (X = 7) = 7 k^{2} + k = \frac{7}{100} + \frac{1}{10} = \frac{17}{100}$

(iv) $P (0 < X < 3) = P (1) + P (2) = 3 k = \frac{3}{10}$

Solved Example 3

Find the mean and variance of

X

with distribution

P (X = 2) = 0.3

P (X = 3) = 0.4

P (X = 4) = 0.3

Solution:

$μ = 2 (0.3) + 3 (0.4) + 4 (0.3) = 0.6 + 1.2 + 1.2 = 3$

$\sum x^{2} p = 4 (0.3) + 9 (0.4) + 16 (0.3) = 1.2 + 3.6 + 4.8 = 9.6$

$σ^{2} = 9.6 - 3^{2} = 0.6$ (Figure 2).

Solved Example 4

A player throws a fair die. He wins Rs 20 if a 6 turns up, wins Rs 10 if a 4 or 5 turns up, and loses Rs 5 otherwise. Find his expected gain per game.

Solution:

Let $X$ be the gain. $X = 20$ with probability $\frac{1}{6}$ , $X = 10$ with probability $\frac{2}{6}$ , $X = - 5$ with probability $\frac{3}{6}$ .

E (X) = 20 \cdot \frac{1}{6} + 10 \cdot \frac{2}{6} - 5 \cdot \frac{3}{6} = \frac{20 + 20 - 15}{6} = \frac{25}{6}

He can expect to gain about Rs 4.17 per game in the long run.

7.2 Binomial probabilities

Solved Example 5

A die is thrown 7 times. What is the probability that an odd number turns up (i) exactly 4 times (ii) at least 4 times?

Solution:

Success = odd number, so $p = \frac{3}{6} = \frac{1}{2}$ and $q = \frac{1}{2}$ , with $n = 7$ .

(i) $P (X = 4) =^{7} C_{4} (\frac{1}{2})^{4} (\frac{1}{2})^{3} = \frac{35}{128}$

(ii) Adding the terms for 4 to 7 successes:

P (X \geq 4) = \frac{^{7} C _{4} + ^{7} C _{5} + ^{7} C _{6} + ^{7} C _{7}}{2 ^{7}} = \frac{35 + 21 + 7 + 1}{128} = \frac{64}{128} = \frac{1}{2}

Figure 5 shows why: the distribution is symmetric.

Solved Example 6

A pair of dice is thrown 4 times. Getting a doublet is a success. Find the probability of exactly two successes.

Solution:

$p = \frac{6}{36} = \frac{1}{6}$ , $q = \frac{5}{6}$ , $n = 4$ .

P (X = 2) =^{4} C_{2} (\frac{1}{6})^{2} (\frac{5}{6})^{2} = 6 \times \frac{25}{1296} = \frac{25}{216}

Solved Example 7

A shooter hits a target with probability

\frac{1}{4}

on each shot, independently. He fires 5 shots. Find the probability that he hits the target at least twice.

Solution:

$n = 5$ , $p = \frac{1}{4}$ , $q = \frac{3}{4}$ . Use the complement of "0 or 1 hit":

P (X \geq 2) = 1 - (\frac{3}{4})^{5} -^{5} C_{1} \cdot \frac{1}{4} (\frac{3}{4})^{4} = 1 - \frac{243}{1024} - \frac{405}{1024} = \frac{376}{1024} = \frac{47}{128}

Solved Example 8

How many times must a fair die be thrown so that the probability of getting at least one six is at least 0.9?

Solution:

P(at least one six in $n$ throws) $= 1 - (\frac{5}{6})^{n} \geq 0.9$ , so $(\frac{5}{6})^{n} \leq 0.1$ .

Taking logarithms: $n \geq \frac{lo g 0.1}{lo g ( 5/6 )} \approx 12.6$ . Check: $(\frac{5}{6})^{12} \approx 0.112$ and $(\frac{5}{6})^{13} \approx 0.093$ .

Answer: at least 13 throws (Figure 7).

Solved Example 9

A fair die is rolled until a six appears. Find the probability that the first six appears on the 4th roll.

Solution:

The first three rolls must fail and the fourth must succeed (Figure 6):

P = q^{3} p = (\frac{5}{6})^{3} \cdot \frac{1}{6} = \frac{125}{1296}

7.3 Mean and variance of the binomial

Solved Example 10

Find the mean and variance of the number of sixes in two throws of a fair die.

Solution:

$X$ = number of sixes, $n = 2$ , $p = \frac{1}{6}$ , $q = \frac{5}{6}$ .

$X$	0	1	2
$P (X)$	$\frac{25}{36}$	$\frac{10}{36}$	$\frac{1}{36}$

Direct: $μ = 0 + \frac{10}{36} + \frac{2}{36} = \frac{1}{3}$ and $\sum x^{2} p = \frac{10}{36} + \frac{4}{36} = \frac{7}{18}$ , so $σ^{2} = \frac{7}{18} - \frac{1}{9} = \frac{5}{18}$ .

Formula check: $n p = \frac{1}{3}$ and $n pq = 2 \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{5}{18}$ .

Solved Example 11

A binomial distribution has mean 8 and variance 4. Find

n

p

and

P (X \leq 2)

Solution:

$n p = 8$ and $n pq = 4$ , so $q = \frac{4}{8} = \frac{1}{2}$ , $p = \frac{1}{2}$ and $n = 16$ .

P (X \leq 2) = \frac{^{16} C _{0} + ^{16} C _{1} + ^{16} C _{2}}{2 ^{16}} = \frac{1 + 16 + 120}{2 ^{16}} = \frac{137}{65536}

Solved Example 12

The difference between the mean and variance of a binomial distribution is 1, and the difference between their squares is 11. Find the probability of exactly 3 successes.

Solution:

$n p - n pq = 1$ and $n^{2} p^{2} - n^{2} p^{2} q^{2} = 11$ .

The first gives $n p (1 - q) = n p^{2} = 1$ . The second gives $n^{2} p^{2} (1 - q) (1 + q) = 11$ . Dividing the second by the square of the first:

\frac{n ^{2} p ^{2} ( 1 - q ) ( 1 + q )}{n ^{2} p ^{2} ( 1 - q ) ^{2}} = \frac{1 + q}{1 - q} = 11 \Rightarrow q = \frac{5}{6}, p = \frac{1}{6}

Then $n p^{2} = 1$ gives $n = 36$ . Check: mean 6, variance 5.

P (X = 3) =^{36} C_{3} (\frac{1}{6})^{3} (\frac{5}{6})^{33}

7.4 JEE-style problems

Solved Example 13

The mean and variance of a binomial distribution are 4 and 2. Then

P (X \geq 1)

is
(A)

1/256

(B)

255/256

(C)

247/256

(D)

9/256

Solution:

$q = \frac{2}{4} = \frac{1}{2}$ , so $p = \frac{1}{2}$ and $n = \frac{4}{1/2} = 8$ .
$P (X \geq 1) = 1 - q^{8} = 1 - \frac{1}{256}$ .

Answer: (B) $\frac{255}{256}$ .

Solved Example 14

A fair coin is tossed 15 times. The most probable number of heads is
(A) 7 only
(B) 8 only
(C) 7 and 8
(D) 7.5

Solution:

$(n + 1) p = 16 \times \frac{1}{2} = 8$ , an integer, so both 8 and $8 - 1 = 7$ are most probable. Check: $^{15} C_{7} =^{15} C_{8} = 6435$ , the largest coefficient.

Answer: (C) 7 and 8. Option (D) is the mean $n p$ , which is not even a possible value here.

Solved Example 15

Three cards are drawn one by one from a pack of 52 with replacement. Find the probability distribution and the mean of the number of aces. How does the answer change without replacement?

Solution:

With replacement, $X \sim B (3, \frac{1}{13})$ : $P (X = k) =^{3} C_{k} (\frac{1}{13})^{k} (\frac{12}{13})^{3 - k}$ .

$X$	0	1	2	3
with replacement	$\frac{1728}{2197}$	$\frac{432}{2197}$	$\frac{36}{2197}$	$\frac{1}{2197}$
without replacement	$\frac{4324}{5525}$	$\frac{1128}{5525}$	$\frac{72}{5525}$	$\frac{1}{5525}$

Mean with replacement $= n p = \frac{3}{13}$ . Without replacement the probabilities are $^{4} C_{k}^{48} C_{3 - k} /^{52} C_{3}$ (not binomial), yet the mean is still $\frac{3}{13}$ , because each draw is an ace with probability $\frac{1}{13}$ on its own.

Solved Example 16

X \sim B (10, 0.4)

. Find

E (3 X + 2)

and

Var (3 X + 2)

Solution:

$E (X) = n p = 4$ and $Var (X) = n pq = 10 \times 0.4 \times 0.6 = 2.4$ .
$E (3 X + 2) = 3 \times 4 + 2 = 14$ .
$Var (3 X + 2) = 3^{2} \times 2.4 = 21.6$ ; the $+ 2$ does not change the spread.

Answer: $14$ and $21.6$ .

Solved Example 17

Find the probability distribution of the number of doublets in three throws of a pair of dice.

Solution:

A doublet has probability $p = \frac{6}{36} = \frac{1}{6}$ , so $X \sim B (3, \frac{1}{6})$ .

$X$	0	1	2	3
$P (X)$	$(\frac{5}{6})^{3} = \frac{125}{216}$	$3 \cdot \frac{1}{6} \cdot \frac{25}{36} = \frac{75}{216}$	$3 \cdot \frac{1}{36} \cdot \frac{5}{6} = \frac{15}{216}$	$\frac{1}{216}$

Check: $125 + 75 + 15 + 1 = 216$ . Mean $= n p = \frac{1}{2}$ .

Practice Questions

A fair coin is tossed 6 times. Find the probability of exactly 4 heads.Answer: $\frac{15}{64}$
$X \sim B (5, p)$ and $P (X = 0) = P (X = 1)$ . Find $p$ .Answer: $\frac{1}{6}$
10% of bulbs from a factory are defective. Find the probability that a random sample of 5 bulbs has no defective bulb.Answer: $(0.9)^{5} = 0.59049$
Find the expected value of the sum of the numbers on two fair dice.Answer: 7
A binomial distribution has mean 6 and variance 2. Find n and p.Answer: $n = 9$ , $p = \frac{2}{3}$
A random variable takes values 1, 2, 3 with probabilities $k$ , $2 k$ , $3 k$ . Find $k$ and the mean.Answer: $k = \frac{1}{6}$ ; mean $= \frac{7}{3}$
Is a binomial distribution with mean 4 and variance 6 possible?Answer: No: $q = \frac{6}{4} > 1$

Common Mistakes to Avoid

Watch out

Using the binomial formula for draws without replacement. The success probability changes, so the trials are not binomial.
Forgetting the $^{n} C_{r}$ factor and writing only $p^{r} q^{n - r}$ , which is the probability of one particular order.
Writing variance $= n p$ . The binomial variance is $n pq$ , and it is always smaller than the mean.
Accepting "mean 4, variance 6" as a binomial distribution. Variance greater than or equal to the mean forces $q \geq 1$ , which is impossible.
Keeping a negative root for $k$ in a distribution table. Every probability must be non-negative.
Writing first success on trial $r$ as $p^{r - 1} q$ . The failures come first: $q^{r - 1} p$ .
Adding many terms for "at least one". Use $1 - q^{n}$ instead.
Writing $Var (a X + b) = a Var (X) + b$ . Constants add nothing to the variance and the scale factor is squared: $a^{2} Var (X)$ .

Frequently Asked Questions

What is a Bernoulli (binomial) trial?

A Bernoulli trial is a single experiment with only two outcomes, success or failure. A sequence of such trials is binomial when the number of trials is fixed, the success probability $p$ stays the same each time, and the trials are independent. Coin tosses and draws with replacement are standard examples.

What is the binomial distribution formula?

If $X$ counts successes in $n$ independent trials with success probability $p$ , then $P (X = r) =^{n} C_{r} p^{r} q^{n - r}$ with $q = 1 - p$ . The $^{n} C_{r}$ counts the orders of the successes, and $p^{r} q^{n - r}$ is the probability of each order.

What are the mean and variance of a binomial distribution?

For $X \sim B (n, p)$ , the mean is $n p$ and the variance is $n pq$ , so the standard deviation is $n pq$ . Because $q$ is less than 1, the variance is always smaller than the mean. Many questions give the mean and variance and ask you to recover $n$ and $p$ .

When can the binomial distribution not be used?

It fails when trials are not independent or the success probability changes, for example when drawing cards or balls without replacement, or when a trial has more than two outcomes of interest. In such cases, count directly with combinations or use the multiplication theorem with conditional probabilities.

How do you find the most probable number of successes?

Compare consecutive terms using $\frac{P ( X = r + 1 )}{P ( X = r )} = \frac{n - r}{r + 1} \cdot \frac{p}{q}$ . The most likely value is the integer part of $(n + 1) p$ . If $(n + 1) p$ is an integer, that value and the one just below it are equally likely.

What is the expected value of a random variable?

The expected value $E (X) = \sum x_{i} p_{i}$ is the long-run average of $X$ over many repetitions, the balance point of its distribution. It need not be a value $X$ can take: a fair die has $E (X) = 3.5$ . For a game, a positive expected gain means you win on average.

Is binomial distribution in the JEE Main 2026 syllabus?

The revised JEE Main syllabus lists "probability distribution of a random variate" but no longer names Bernoulli trials and the binomial distribution separately. Repeated independent trial questions can still be solved with the multiplication theorem and combinations, so the binomial formula remains a useful and time-saving tool.

Does JEE Advanced include mean and variance of a random variable?

Yes. The JEE Advanced 2026 syllabus lists random variables along with their mean and variance, next to measures of dispersion in statistics. Expect questions that give a probability table with an unknown constant, ask for the expected value of a game, or combine a distribution with counting.

Previous year questions on Binomial Trials And Binomial Distribution

5 questions from past papers, each with a step-by-step solution.

Ready to master Probability?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test